If the function f:R→R defined by f(x)=5−3x, if x≤53x2−3x+20, if x>53then 'f' is

limx→5−3f(x)=5−3(53)=0 limx→5+3f(x)=limx→5+3x2−3x+20 =(53)2−3(53)+20=1609limx→5−3f(x)≠limx→5+3f(x)f(x) is discontinuous x=53Hence f(x) is devivative at x = 2

If the function f:R→R defined by f(x)=5−3x, if x≤53x2−3x+20, if x>53then 'f' is

limx→5−3f(x)=5−3(53)=0 limx→5+3f(x)=limx→5+3x2−3x+20 =(53)2−3(53)+20=1609limx→5−3f(x)≠limx→5+3f(x)f(x) is discontinuous x=53Hence f(x) is devivative at x = 2

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If the function $f : R \to R$ defined by $f (x) = \{\begin{cases} 5 - 3 x, i f x \leq \frac{5}{3} \\ x^{2} - 3 x + 20, i f x > \frac{5}{3} \end{cases}$ then 'f' is

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discontinuous at $x = - 2$

continuous at $x = \frac{5}{3}$

differentiable at $x = \frac{5}{3}$

differentiable at $x = 2$

answer is C.

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Detailed Solution

$\lim_{x \to \frac{5^{-}}{3}} f (x) = 5 - 3 (\frac{5}{3}) = 0 \lim_{x \to \frac{5^{+}}{3}} f (x) = \lim_{x \to \frac{5^{+}}{3}} x^{2} - 3 x + 20 \begin{array}{l} = {(\frac{5}{3})}^{2} - 3 (\frac{5}{3}) + 20 = \frac{160}{9} \\ \lim_{x \to \frac{5^{-}}{3}} f (x) \neq \lim_{x \to \frac{5^{+}}{3}} f (x) \end{array}$
f(x) is discontinuous $x = \frac{5}{3}$
Hence f(x) is devivative at x = 2